Quotient Theorem for Group Homomorphisms/Examples/Real to Complex Numbers under e^2 pi i x

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Example of Use of Quotient Theorem for Group Homomorphisms

Let $\struct {\R, +}$ denote the additive group of real numbers.

Let $\struct {\C_{\ne 0}, \times}$ denote the multiplicative group of complex numbers.

Let $\phi: \struct {\R, +} \to \struct {\C_{\ne 0}, \times}$ be the homomorphism defined as:

$\forall x \in \R: \map \phi x = e^{2 \pi i x}$


Then $\phi$ can be decomposed into the form:

$\phi = \alpha \beta \gamma$

in the following way:


$\alpha: \struct {K, \times} \to \struct {\C_{\ne 0}, \times}$ is defined as:
$\forall z \in K: \map \alpha z = z$
where $\struct {K, \times}$ denotes the circle group:
$K = \set {z \in \C: \cmod z = 1}$
$\times$ is the operation of complex multiplication


$\beta: \hointr 0 1 \to K$ is defined as:
$\forall x \in \hointr 0 1: \map \beta x = e^{2 \pi i x}$
where $\hointr 0 1$ denotes the right half-open real interval $\set {x \in \R: 0 \le x < 1}$


$\gamma: \R \to \hointr 0 1$ is defined as:
$\forall x \in \R: \map \gamma x = \fractpart x$
where $\fractpart x$ is the fractional part of $x$:
$\fractpart x := x - \floor x$


Proof

It is first demonstrated that $\phi$ is a homomorphism:

\(\ds \map \phi {x + y}\) \(=\) \(\ds e^{2 \pi i \paren {x + y} }\) Definition of $\phi$
\(\ds \) \(=\) \(\ds e^{2 \pi i x + 2 \pi i y}\)
\(\ds \) \(=\) \(\ds e^{2 \pi i x} e^{2 \pi i y}\)
\(\ds \) \(=\) \(\ds \map \phi x \, \map \phi y\)


We have that:

\(\ds \map \phi 0\) \(=\) \(\ds e^{2 \pi i 0}\)
\(\ds \) \(=\) \(\ds e^0\)
\(\ds \) \(=\) \(\ds 1\)

By Group Homomorphism Preserves Identity it is confirmed that $1$ is the identity of $\struct {\C_{\ne 0}, \times}$.


Now we can establish what the kernel of $\phi$ is:

\(\ds \map \phi x\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds e^{2 \pi i x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \cos 2 \pi x + i \sin 2 \pi x\) \(=\) \(\ds 1\) Euler's Formula
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \Z\) Cosine of Integer Multiple of Pi and Sine of Integer Multiple of Pi

Thus:

$\map \ker \phi = \Z$

where $\Z$ denotes the set of integers.


Next we establish what the image of $\phi$ is:

\(\ds z\) \(\in\) \(\ds \Img \phi\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in \R: \, \) \(\ds z\) \(=\) \(\ds e^{2 \pi i x}\)
\(\ds \leadsto \ \ \) \(\ds \ln z\) \(=\) \(\ds 2 \pi i x\) Definition 2 of Complex Logarithm
\(\ds \leadsto \ \ \) \(\ds \ln \cmod z + i \arg z\) \(=\) \(\ds 2 \pi i x\) Definition 1 of Complex Logarithm
\(\ds \leadsto \ \ \) \(\ds \ln \cmod z\) \(=\) \(\ds 0\) separating into real and imaginary parts
\(\ds \arg z\) \(=\) \(\ds 2 \pi x\)
\(\ds \leadsto \ \ \) \(\ds \cmod z\) \(=\) \(\ds 1\) Natural Logarithm of 1 is 0
\(\ds \Img {\arg z}\) \(=\) \(\ds \R\)
\(\ds \leadsto \ \ \) \(\ds \Img \phi\) \(=\) \(\ds \set {z \in \C_{\ne 0}: \cmod z = 1}\)
\(\ds \) \(=\) \(\ds K\) Definition of Circle Group


Thus, from the Quotient Theorem for Group Homomorphisms, $\phi$ can be decomposed into:

$\phi = \alpha \beta \gamma$

where:

$\alpha: K \to \C_{\ne 0}$, which is a monomorphism
$\beta: \R / \Z \to K$, which is an isomorphism
$\gamma: \R \to \R / \Z$, which is an epimorphism.


As $\beta$ is an isomorphism, $\beta$ is also a bijection and so $\R / \Z = \Preimg \beta$ can be deduced:

\(\ds x\) \(\in\) \(\ds \Preimg \beta\)
\(\ds \leadsto \ \ \) \(\ds \exists z \in K: \, \) \(\ds z\) \(=\) \(\ds e^{2 \pi i x}\)
\(\ds \leadsto \ \ \) \(\ds \Ln z\) \(=\) \(\ds \ln r + i \theta\) for $\theta \in \hointr 0 {2 \pi}$ \(\quad\) Definition of Principal Branch of Complex Natural Logarithm

We have specifically selected $\hointr 0 {2 \pi}$ as the image of the principal argument of $\Ln z$.

Other half-open real intervals whose length is $2 \pi$ work equally well, for example $\hointl {-\pi} \pi$.


Thus:

\(\ds x\) \(\in\) \(\ds \Preimg \beta\)
\(\ds \leadsto \ \ \) \(\ds \Arg {\map \beta x}\) \(=\) \(\ds 2 \pi x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \hointr 0 1\)

Thus we have:

$\Preimg \beta = \R / \Z = \hointr 0 1$

and the result follows.

$\blacksquare$


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